Nilpotent Element is Contained in Prime Ideals

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Let $R$ be a commutative ring with unity.

Let $\mathfrak p \subset R$ be a prime ideal of $R$.

Let $x \in R$ be a nilpotent element of $R$.

Then $x \in \mathfrak p$.


Let $x$ be nilpotent in $R$ as asserted.

Then by definition:

$\exists n \in \Z_{>0}: x^n = 0$

But $0 \in \mathfrak p$ so:

$x^n \in \mathfrak p$

\(\ds x\) \(\in\) \(\ds \map \Rad {\mathfrak p}\) Definition 1 of Radical of Ideal of Ring
\(\ds \) \(=\) \(\ds \mathfrak p\) Radical of Prime Ideal